Ignorant Bidding

The other day at work, the great Hungarian polymath John von Neumann came up in a conversation. Enumerating some of his many contributions to math and science, I said that he was one of the founders of game theory, having coauthored Theory of Games and Economic Behavior with Oskar Morgenstern. The gentleman to whom I was speaking replied that he had used game theory in his MBA program, and while he said that, I thought of a problem set exercise from my freshman-year probability course. Now I am writing a blog post about this problem. It’s both strange and fascinating how the mind works sometimes.

The problem, which can be found in its original form here on Joe Blitzstein’s Stat 110 course website, concerns how much one should bid for an asset (e.g. a painting, a car, real estate, etc.) with value unknown to the bidder. Not only is the exact value unknown, but the bidder has no information on the asset’s value V, except that it is at least $0 and at most $1 million. Therefore we treat the asset’s value as a random variable V uniformly distributed between 0 and 1 (in $ million). Even though the bidder is ignorant of the asset’s value, he or she knows the conditions under which the bid b will be accepted or rejected. If b<\frac{2}{3}V, then the bid is rejected; if b\geq\frac{2}{3}V, then the bid is accepted. Given this information, how much should one bid to maximize his or her expected payoff, the value of the asset minus the bid?

In tackling this problem, it’s important to understand that the bidder’s payoff X is a function of the random variable V. Indeed,

\displaystyle X=\begin{cases}0&{b<\frac{2}{3}V}\\{V-b}&{b\geq\frac{2}{3}V}\end{cases}

To compute the expected payoff as function of the bid b, we use conditional expectation. Observe that

\displaystyle\mathbb{E}[X\mid V]=(V-b)\mathbf{1}_{b\geq\frac{2}{3}V}

By the tower property of conditional expectation, we see that

\begin{array}{lcl}\displaystyle\mathbb{E}_{b}[X]=\mathbb{E}_{b}\left[\mathbb{E}[X\mid V]\right]=\int_{V\leq\frac{3}{2}b}(V-b)d\mathbb{P}&=&\displaystyle\int_{0}^{\frac{3}{2}b}(v-b)dv\\[1 em]&=&\displaystyle\left[\dfrac{1}{2}v^{2}-bv\right]_{v=0}^{v=\frac{3}{2}b}\\[1.2 em]&=&\displaystyle\left(\dfrac{1}{2}\cdot\dfrac{9b^{2}}{4}-\dfrac{3b^{2}}{2}\right)\\[.9 em]&=&\displaystyle-\dfrac{3b^{2}}{8}\end{array}

Our computation shows that, regardless of how high or low one bids for the asset, the expected payoff is nonpositive. The expected payoff is maximized by bidding $0. In other words, don’t bid for an asset without any information of its value!

Actually, that last assertion isn’t quite true. If we change the scaling factor of V from \frac{2}{3} to some c\in(0,1) in determining whether the bid will be accepted or rejected, then the expected payoff is not necessarily maximized by bidding $0. Indeed, the above calculation becomes

\begin{array}{lcl}\displaystyle\mathbb{E}_{b}[X]=\int_{V\leq\min\left\{\frac{b}{c},1\right\}}(V-b)d\mathbb{P}=\int_{0}^{\min\left\{\frac{b}{c},1\right\}}(v-b)dv&=&\displaystyle\left[\dfrac{1}{2}v^{2}-bv\right]_{v=0}^{v=\min\left\{\frac{b}{c},1\right\}}\\[1.1 em]&=&\displaystyle\dfrac{1}{2}{\min\left\{\frac{b}{c},1\right\}^{2}}-b\cdot{\min\left\{\frac{b}{c},1\right\}}\end{array}

If \frac{b}{c}>1, then

\displaystyle\mathbb{E}_{b}[X]=\dfrac{1}{2}-b\geq0\Longleftrightarrow\dfrac{1}{2}\geq b,

which implies that b must satisfy the inequality c<b\leq\frac{1}{2}. If \frac{b}{c}\leq1, then

\displaystyle\mathbb{E}_{b}[X]=\dfrac{b^{2}}{2c^{2}}-\dfrac{b^{2}}{c}\geq0\Longleftrightarrow\dfrac{1}{2}\geq c

So if c>\frac{1}{2}, then the expected payoff is maximized by bidding $ 0 or rather, not bidding. If c<\frac{1}{2}, then the expected payoff is maximized by $ c. If c=\frac{1}{2}, then the expected payoff is maximized (i.e. zero) by bidding $ c or less.

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